Question:

A parallel plate capacitor is constructed using two square plates, each with side length L, and separated by a distance d. The space between the plates is filled with a dielectric material of relative permittivity, ε_r. The capacitor is connected to a battery of voltage V, causing a uniform electric field to form between the plates.

1. Derive an expression for the capacitance, C, of this parallel plate capacitor in terms of the given variables.
2. A new capacitor is formed by doubling the distance between the plates, while keeping the other parameters constant. Determine the effect of this change on the capacitance, electric field, charge stored, energy stored, and potential energy stored in the new capacitor. Justify your reasoning using relevant equations.

Answer:

1. To derive an expression for the capacitance of a parallel plate capacitor, we can start by considering the electric field between the plates. The electric field, E, between the plates is given by:

   E = V/d

   Next, we can relate the electric field to the charge density, σ, on the plates. The charge density, σ, is defined as the charge per unit area. Since the plates are square, the area A of each plate is given by:

   A = L^2

   The charge density, σ, is related to the electric field, E, as:

   σ = ε₀ε_rE

   where ε₀ is the vacuum permittivity. Substituting for E, we get:

   σ = ε₀ε_r(V/d)

   The total charge, Q, on one plate is given by:

   Q = σA

   Substituting for σ and A, we get:

   Q = ε₀ε_r(V/d)(L^2)

   The capacitance, C, is defined as the ratio of the charge, Q, to the voltage, V:

   C = Q/V

   Substituting for Q, we get the expression for the capacitance, C:

   

   Therefore, the capacitance of the parallel plate capacitor is given by:

   

2. When the distance between the plates is doubled, while keeping the other parameters constant, the capacitance, electric field, charge stored, energy stored, and potential energy stored in the new capacitor will change as follows:

   • Capacitance (C): According to the expression derived in Part 1, the capacitance is inversely proportional to the distance (d). Doubling the distance will result in halving the capacitance.

   • Electric Field (E): The electric field is directly proportional to the voltage (V) and inversely proportional to the distance (d). If the voltage remains constant and the distance is doubled, the electric field will decrease by a factor of 2.

   • Charge Stored (Q): The charge stored on the plates is directly proportional to the capacitance (C) and the voltage (V). If the capacitance is halved, the charge stored will also be halved.

   • Energy Stored (U): The energy stored in the capacitor is given by the equation: U = (1/2)CV^2. Since the capacitance is halved and the voltage remains constant, the energy stored will be reduced by a factor of 4.

   • Potential Energy Stored (PE): The potential energy stored in the capacitor is given by the equation: PE = (1/2)QV. As the charge stored (Q) is halved and the voltage remains constant, the potential energy stored will also be halved.

Therefore, doubling the distance between the plates in a parallel plate capacitor, while keeping other parameters constant, results in halving the capacitance (C), electric field (E), charge stored (Q), potential energy (PE) stored, and reducing the energy stored (U) by a factor of 4.